Opuscula Mathematica •Vol. 29•No. 4 •2009
Janusz Matkowski
EXTENSIONS OF SOLUTIONS
OF A FUNCTIONAL EQUATION IN TWO VARIABLES
Abstract. An extension theorem for the functional equation of several variables
f(M(x, y)) =N(f(x), f(y)),
where the given functionsM andN are left-side autodistributive, is presented.
Keywords: functional equation, autodistributivity, strict mean, extension theorem.
Mathematics Subject Classification: Primary 39B22.
1. AN EXTENSION THEOREM
Recently Zs. Páles [2] considered the extension problem for the functional equation of the form
f(x) =M(f(m1(x, y)), . . . , f(mk(x, y)))
whereM is ak-variable bisymmetric operation andm1, . . . , mk some binary commut-ing operations.
In this note we deal with the extension problem for the functional equation
f(M(x, y)) =N(f(x), f(y))
where the given functions M and N are left-side autodistributive and M is a strict mean. The functions satisfying this functional equation are called(M, N)-affine (cf. [1] where the case when bothM andN are means is considered).
We prove the following
Theorem 1.1. LetI, J ⊂Rbe open intervals. Suppose thatM :I2→I,N :J2→J are continuous strictly increasing with respect to the first variable and such that
M(M(x, y), z)) =M(M(x, z), M(y, z)), x, y, z∈I, (1.1)
424 Janusz Matkowski
N(N(x, y), z)) =N(N(x, z), N(y, z)), x, y, z∈J. (1.2)
We also assume thatM is a strict mean, that is
min(x, y)< M(x, y)<max(x, y), x, y∈I, x6=y.
If f : I0→J satisfies the functional equation
f(M(x, y)) =N(f(x), f(y)), x, y∈I0, (1.3)
for a nontrivial intervalI0⊂I, then there exists a unique function F :I→J such that F|I0 =f and
F(M(x, y)) =N(F(x), F(y)), x, y∈I.
Proof. Assume thatI0⊂Iis a maximal subinterval ofIon which the functionf can be extended to satisfy equation (1.3). Suppose first that
b:= supI0<supI.
Take an arbitrarya∈I0,a < b. Then
f(M(x, y)) =N(f(x), f(y)), x, y∈[a, b]. (1.4)
SinceM is a continuous and strict mean, there is ac∈I,c > bsuch thatM(a, c)< b. Hence, asM is strictly increasing with respect to the first variable,
M(x, c)< b, x∈[a, c].
Setting y:=ain (1.4) we have
f(M(x, a)) =N(f(x), f(a)), x∈[a, c].
LetN−1
f(a)denote the inverse function ofN(·, f(a)). DefineFa : [a, c]→J by
Fa(x) :=N
−1
f(a)(f(M(x, a))), x∈[a, c].
Note that by (1.4) the functionFa is correctly defined,
Fa(x) =f(x), x∈[a, b],
and
f(M(x, a)) =N(Fa(x), f(a)), x∈[a, c]. (1.5)
Now making use respectively of the definition ofFa, the property (1.1) ofM, equation (1.4), equation (1.5), property (1.2) ofN, for allx, y∈[a, c]we have
Fa(M(x, y)) :=N
−1
f(a)(f(M(M(x, y), a))) =N
−1
f(a)(f(M(M(x, a), M(y, a))) =
=N−1
f(a)(N(f(M(x, a)), f(M(y, a)))) =
=N−1
f(a)(N(N(Fa(x), f(a)), N(Fa(y), f(a)))) =
=N−1
Extensions of solutions of a functional equation in two variables 425
that is
Fa(M(x, y)) =N(Fa(x), Fa(y)), x, y∈[a, c]. (1.6)
In the case when infI0∈I0 we can takea= infI0. PuttingF :=Fa we get
F(M(x, y)) =N(F(x), F(y)), x, y∈[a, c].
SinceF|I0 =f andI0 [a, c], this contradicts the maximality of the interval I0. In the case when infI0∈/I, we take a decreasing sequence(an:n∈N)such that
infI0= limn→∞an and defineF : (infI0, c]→J by
F(x) :=Fan(x), x∈[an, c], n∈N.
This definition is correct because, by (1.3),
m < n=⇒Fam =Fan
[am,c].
In view of (1.6), we have
F(M(x, y)) =N(F(x), F(y)), x, y∈(infI0, c].
SinceF|I0 =f andI0 (infI0, c], this contradicts the maximality of the intervalI0. The obtained contradiction proves that supI0 = supI. In a similar way we can show thatinfI0= infI. This completes the proof.
REFERENCES
[1] J. Matkowski,Convex functions with respect to a mean and a characterization a quasi--arithmetic mean, Real Anal. Exchange29(2003/2004) 1, 220–246.
[2] Zs. Páles, Extension theorems for functional equations with bisymmetric operations, Aequationes Math.63(2002), 266–291.
Janusz Matkowski
j.matkowski@wmie.im.zgora.pl
University of Zielona Góra
Institute of Mathematics, Computer Science and Econometry, ul. Podgórna 50, 65-246 Zielona Góra, Poland
Silesian University Institute of Mathematics 40-007 Katowice, Poland
